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A stationary body of mass 3 kg explodes ...

A stationary body of mass 3 kg explodes into three equal pieces. Two of the pieces fly off in two mutually perpendicular directions, one with a velocity of `3hati ms^(1)` and the other with a velocity of `4hatj ms^(-1)` . - If the explosion occurs in `10^(–4) s`, the average force acting on the third piece in newton is

A

`(3hat(i) + 4hat(j)) xx 10^(-4)`

B

`(3hat(i) - 4hat(j)) xx 10^(-4)`

C

`(3hat(i) + 4hat(j)) xx 10^(4)`

D

`-(3hat(i) + 4hat(j)) xx 10^(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
Law of conservation of momentum
`m_(3)v_(3) = - 1 xx (3hat(i) + 4hat(j)) kg ms^(-1)`
Impulse = Average force `xx` Time
`rArr` Average force ` = ("Impulse")/("Time") = ("Change in momentum")/("Time")`
` = (-(3hat(i) + 4hat(j)))/(10^(-4)) = - (3 hat(i ) + 4 hat(j)) xx 10^(4) N`
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